Answer:
Ix = Iy =
Radius of gyration x = y =
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is . Moment of inertia of quarter disk about its perpendicular is
.
Now using perpendicular axis theorem, Ix = Iy = =
.
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.
Mr. Mole's burrow was at an altitude of 6 meters below the ground.
Step-by-step explanation:
Step 1:
We need to determine the distance that Mr. Mole covers in a single minute.
To do that we divide the difference in values of altitude by the difference in the time periods.
For the first case, Mr. Mole had traveled -18 meters in 5 minutes.
We also have, he traveled -25.2 meters in 8 minutes.
Step 2:
The distance he covered in 1 minute
So with every minute, Mr. Mole digs down an additional 2.4 meters below the surface.
To determine where Mr. Mole's burrow is we subtract the distance traveled in 5 minutes from -18.
The altitude of Mr. Mole's burrow
So Mr. Mole's burrow was at an altitude of 6 meters below the ground i.e. -6 meters.